Theorem sbc3or 44983

Description: sbcor 3780 with a 3-disjuncts. This proof is sbc3orgVD 45301 automatically translated and minimized. (Contributed by Alan Sare, 31-Dec-2011.) (Revised by NM, 24-Aug-2018.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
sbc3or	⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓 ∨ [𝐴 / 𝑥]𝜒))

Proof of Theorem sbc3or

Step	Hyp	Ref	Expression
1		sbcor 3780	. . 3 ⊢ ([𝐴 / 𝑥]((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ∨ [𝐴 / 𝑥]𝜒))
2		df-3or 1093	. . . . 5 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∨ 𝜒))
3	2	bicomi 225	. . . 4 ⊢ (((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ (𝜑 ∨ 𝜓 ∨ 𝜒))
4	3	sbcbii 3786	. . 3 ⊢ ([𝐴 / 𝑥]((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ [𝐴 / 𝑥](𝜑 ∨ 𝜓 ∨ 𝜒))
5		sbcor 3780	. . . 4 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓))
6	5	orbi1i 919	. . 3 ⊢ (([𝐴 / 𝑥](𝜑 ∨ 𝜓) ∨ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
7	1, 4, 6	3bitr3i 302	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
8		df-3or 1093	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓 ∨ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
9	7, 8	bitr4i 279	1 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓 ∨ [𝐴 / 𝑥]𝜒))

Syntax hints:   ↔ wb 207   ∨ wo 853   ∨ w3o 1091  [wsbc 3730

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-10 2152  ax-12 2189  ax-ext 2712

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-3or 1093  df-tru 1550  df-ex 1787  df-nf 1791  df-sb 2074  df-clab 2719  df-cleq 2732  df-clel 2815  df-v 3434  df-sbc 3731

This theorem is referenced by:  sbcoreleleq  44986